Đáp án B

a) 3/4 - x = 1                                  c) x - 1/5 = 2

            x = 3/4 - 1                              x          = 2 + 1/5

            x = -1/4                                  x          = 11/5

Vậy x = -1/4                                   Vậy x = 11/5                                

b) x+4 = 1/5                                  d) x + 5/3 = 1/81

    x      = 1/5 -4                                 x           = 1/81 -5/3

    x     =  -19/5                                  x           = -134/81

Vậy x = -19/5

a) 3/4 - x = 1

\(x=\frac{3}{4}-1\)

\(x=\frac{-1}{4}\)

b) x + 4 = 1/5

\(x=\frac{1}{5}-4\)

\(x=\frac{-19}{5}\)

c) x - 1/5 = 2

\(x=2+\frac{1}{5}\)

\(x=\frac{11}{5}\)

d) x + 5/3 = 1/81

\(x=\frac{1}{81}-\frac{5}{3}\)

\(x=\frac{-134}{81}\)

chúc bạn học tốt

a: Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

b: Thay \(x=\dfrac{1}{4}\) vào P, ta được:

\(P=\left(\dfrac{1}{2}-1\right):\left(\dfrac{1}{2}+1\right)=\dfrac{-1}{2}:\dfrac{3}{2}=-\dfrac{1}{3}\)

c: Ta có: \(P< \dfrac{1}{2}\)

\(\Leftrightarrow P-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

lớp 6 gì kinh thế cái này lớp 8

M=a^3+b^3+ab

M=(a+b)[(a+b)^2-3ab)]+ab=1-2ab 

a+b=1=> b=1-a

M=1-2a(1-a)=1+2a^2-2a

M=2.[(a^2-a+1/2)]+1

-=2(a-1/2)^2+1/2

GTLN của M=1/2 khi a=b=1/2

a, \(x=\dfrac{3}{4}-1=\dfrac{3}{4}-\dfrac{4}{4}=-\dfrac{1}{4}\)

b, \(x=\dfrac{1}{5}-4=\dfrac{1}{5}-\dfrac{20}{5}=-\dfrac{19}{5}\)

c, \(x=2+\dfrac{1}{5}=\dfrac{10}{5}+\dfrac{1}{5}=\dfrac{11}{5}\)

d, \(x=\dfrac{1}{81}-\dfrac{5}{3}=\dfrac{1}{81}-\dfrac{135}{81}=-\dfrac{134}{81}\)

\(a,\dfrac{3}{4}-x=1\\ \Rightarrow x=1,75\\ b,x+4=\dfrac{1}{5}\\ \Rightarrow x=-3,8\\ c,x-\dfrac{1}{5}=2\\ \Rightarrow x=2,2\\ d,x+\dfrac{5}{3}=\dfrac{1}{81}\\x=\dfrac{1}{81}-\dfrac{5}{3}\\ \Rightarrow x=-\dfrac{134}{81}\)

a) \(\dfrac{3}{4}-x=1\)

\(x=\dfrac{3}{4}-1\)

\(x=-\dfrac{1}{4}\)\(=0,25\)

b) \(x+4=\dfrac{1}{5}\)

\(x=\dfrac{1}{5}-4\)

\(x=-\dfrac{19}{5}\)\(=-3,8\)

c) \(x-\dfrac{1}{5}=2\)

\(x=2+\dfrac{1}{5}\)

\(x=\dfrac{11}{5}=2,2\)

d) \(x+\dfrac{5}{3}=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}-\dfrac{5}{3}\)

\(x=-\dfrac{134}{81}\)

a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)

\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)

\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)

\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)

\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)

b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)

\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)

\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)

\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)

\(\Leftrightarrow x=\frac{1}{-3}\)

c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)

\(\Leftrightarrow x=\frac{5}{6}\)

d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)

\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{6}\)

a: x=15

c: \(\Leftrightarrow2x+1=3\)

hay x=1

a) \(\Rightarrow x=2021-2006=15\)

b) \(\Rightarrow2x-2016=64\Rightarrow2x=2016+64=2080\Rightarrow x=1040\)

c) \(\Rightarrow\left(2x+1\right)^3=81:3=27\Rightarrow2x+1=3\)

\(\Rightarrow2x=3-1=2\Rightarrow x=1\)

`a) x= 15`

`b) 2x-2016=16\(\times\)4`

`⇔ 2x-2016=64`

`⇔ 2x= 2080`

`⇔ x= 1040`

`c) 3(2x+1)^3 =81`

`⇔ (2x+1)^3 =27`

`⇔ (2x+1)^3 =3^3`

`⇔ 2x+1=3`

`⇔ x= 1`

a) \(\left(x-1\right)^3\)

\(=x^3-3x^2+3x-1\)

b) \(\left(2x-3y\right)^3\)

\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)

\(=8x^3-36x^2y+54xy^2-27y^3\)

Bài 3: 

a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)

\(\Leftrightarrow12x=13\)

hay \(x=\dfrac{13}{12}\)

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)

\(\Leftrightarrow x^3-1-x^3+4x=4\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)